A liquid freezes at and boils at . If the and values for the liquid are 5.6 and 2.5 deg/molal, respe — Solutions and Colligative Properties Chemistry Question
Question
A liquid freezes at $7^\circ\text{C}$ and boils at $77^\circ\text{C}$. If the $K_f$ and $K_b$ values for the liquid are 5.6 and 2.5 deg/molal, respectively, then the ratio of molar latent heat of fusion to molar latent heat of vaporization is
💡 Solution & Explanation
Cryoscopic and ebullioscopic constants are given by $K = (RT^2M) / (1000\Delta H)$. Hence, $K_f / K_b = (T_f^2 / \Delta H_{\text{fus}}) \times (\Delta H_{\text{vap}} / T_b^2)$. We have $5.6 / 2.5 = ((273+7)^2 / (273+77)^2) \times (\Delta H_{\text{vap}} / \Delta H_{\text{fus}})$. $5.6 / 2.5 = (280/350)^2 \times (\Delta H_{\text{vap}} / \Delta H_{\text{fus}}) = (4/5)^2 \times (\Delta H_{\text{vap}} / \Delta H_{\text{fus}}) = (16/25) \times (\Delta H_{\text{vap}} / \Delta H_{\text{fus}})$. Thus, $\Delta H_{\text{fus}} / \Delta H_{\text{vap}} = (16/25) \times (2.5/5.6) = (16/25) \times (25/56) = 16/56 = 2/7$. Therefore, correct answer is C.